Question function spaces

Question

Let $I\subseteq \mathbb{R}$ be an open interval and define $C^{n}_{b}(I):=\{f \in C(I): f \text{ n-times cont. differentiable and } \lvert \lvert f \rvert \rvert_{n, \infty} < \infty\}$, where $\lvert \lvert f \rvert \rvert_{n, \infty}:=\sum_{i=0}^{n}\lvert \lvert f^{(i)} \rvert \rvert_{\infty}$. Show that $$a)\text{ }(C^{1}_{b}(I),\lvert \lvert f \rvert \rvert_{1, \infty}) \text{ is complete} \\ b) \text{ } (C^{n}_{b}(I),\lvert \lvert f \rvert \rvert_{\infty}) \text{ is not complete for } n \geq 2.$$ Assume w.l.o.g. that $0 \in I$. My approach for b) is to define $f_{m}(x):=(x^{2}+\frac{1}{m})^{n-\frac{1}{2}}$. Then $f_{m} \in C^{n}_{b}(I)$ and $f_{m} \rightarrow f$, whose is defined as $f(x):=(x^{2})^{n-\frac{1}{2}}$, which n-th derivative doesn't exist at $0$. Is that right? Concerning a) im clueless...could anyone help?

Answer 1

I'm missing something - they all look complete. (Oh: I read it wrong. Part (b) does not say that $(C^n_b,||\cdot||_{C^n_b}$ is not complete!)

Calculus Lemma If $f_n$ is differentiable, $f_n'\to g$ uniformly and $f_n\to f$ pointwise then $f$ is differentiable, with $f'=g$.

(We only need that for continuously differentiable functions, in which case it's clear from the Fundamental Theorem of Calculus.)

Now say $(f_n)$ is Cauchy in $C^2_b$. In particular, all three sequences $(f_n), (f'_n)$ and $(f''_n)$ are uniformly Cauchy. So we have $f_n\to f$ uniformly, $f_n'\to g$ uniformly and $f_n''\to h$ uniformly. Now the lemma shows that $g=f'$ and $h=g'=f''$, so $f\in C^2_b$, and in fact $f_n\to f$ in the $C^2_b$ norm. So $C^2_b$ is complete.